Tensor Product vs. Cartesian Product in Quantum Systems

For a composite quantum system, the standard construction is the tensor product of the Hilbert spaces for its parts. A Cartesian product is useful for pairing labels or recording two subsystem states, but it does not provide the general state space of the combined quantum system.

This difference matters because a joint system can be in a superposition of combinations of subsystem states. It can also be entangled, meaning its state cannot be described as a state for each part separately.

The two constructions

Suppose subsystem A has Hilbert space $\mathcal{H}_A$ and subsystem B has Hilbert space $\mathcal{H}_B$.

The Cartesian product $\mathcal{H}_A \times \mathcal{H}_B$ consists of ordered pairs $(|\psi_A\rangle, |\psi_B\rangle)$. If treated as a vector space, its operations act component by component. For finite-dimensional spaces, its dimension is the sum of the dimensions of its factors.

The tensor product $\mathcal{H}_A \otimes \mathcal{H}_B$ is built to represent joint quantum states. If the subsystem dimensions are $m$ and $n$, its dimension is $mn$. A tensor product of subsystem vectors, written $|\psi_A\rangle \otimes |\psi_B\rangle$, is one particular kind of joint state: a product state.

In physics, the word “direct product” is sometimes used informally for a Cartesian product. It can also refer to other constructions, so the intended meaning should be checked in context. The relevant quantum-mechanical construction for distinguishable subsystems is the tensor product.

How tensor-product basis states work

Choose bases $\{|i\rangle_A\}$ and $\{|j\rangle_B\}$ for the two subsystems. The joint basis consists of states

$$|i\rangle_A \otimes |j\rangle_B,$$

one for every pair of basis labels $(i,j)$. Thus, the labels for joint basis states come from a Cartesian product of label sets, while the vectors themselves belong to a tensor-product space.

For example, two two-level systems each have a two-dimensional Hilbert space. Their joint space has dimension $2 \times 2 = 4$, with basis $|00\rangle, |01\rangle, |10\rangle, |11\rangle$, where $|01\rangle$ abbreviates $|0\rangle_A \otimes |1\rangle_B$.

Worked example: two qubits

Let the state of qubit A be

$$|\psi_A\rangle = \alpha|0\rangle + \beta|1\rangle,$$

and the state of qubit B be

$$|\psi_B\rangle = \gamma|0\rangle + \delta|1\rangle,$$

where each state is normalized: $|\alpha|^2+|\beta|^2=1$ and $|\gamma|^2+|\delta|^2=1$. Their joint product state is

$$\begin{aligned} |\psi_A\rangle \otimes |\psi_B\rangle &= (\alpha|0\rangle + \beta|1\rangle) \otimes (\gamma|0\rangle + \delta|1\rangle) \\ &= \alpha\gamma|00\rangle + \alpha\delta|01\rangle + \beta\gamma|10\rangle + \beta\delta|11\rangle. \end{aligned}$$

The pair $(|\psi_A\rangle, |\psi_B\rangle)$ records the separate subsystem states. The tensor product combines their amplitudes into a vector in the four-dimensional joint space. In general, a joint state is an arbitrary normalized vector in that space, not just one formed from a pair of subsystem vectors.

Why the tensor product includes entanglement

A familiar joint state is the Bell state

$$|\Phi^+\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}.$$

It cannot be written as $|\psi_A\rangle \otimes |\psi_B\rangle$. To see why, suppose each qubit were in a separate state, $(a|0\rangle+b|1\rangle) \otimes (c|0\rangle+d|1\rangle)$. Its amplitudes for $|00\rangle, |01\rangle, |10\rangle, |11\rangle$ are $ac, ad, bc, bd$. Matching the Bell state requires nonzero amplitudes for $|00\rangle$ and $|11\rangle$, but zero amplitudes for $|01\rangle$ and $|10\rangle$. Those conditions cannot all hold for a product of two single-qubit states. The Bell state is therefore entangled.

A Cartesian pair of single-qubit states cannot capture this state: it specifies one state per qubit, whereas an entangled state belongs to the joint tensor-product space and has no such factorization.

Common mistakes

• Confusing pairs of labels with pairs of state vectors. The outcome labels $(0,1)$ can index the joint basis state $|01\rangle$. That does not make the full joint state space a Cartesian product of subsystem state vectors.
• Assuming every joint state is a product state. Product states are a subset of the joint space. Superpositions such as Bell states are valid joint states but are not products.
• Adding dimensions instead of multiplying them. For finite-dimensional subsystem Hilbert spaces, $\dim(\mathcal{H}_A \otimes \mathcal{H}_B)=\dim(\mathcal{H}_A)\dim(\mathcal{H}_B)$. The Cartesian product of the underlying vector spaces, with componentwise operations, instead has dimension equal to the sum.
• Forgetting the identical-particle qualification. For distinguishable subsystems, the full tensor product is the standard joint space. For identical particles, physical states must also obey the appropriate exchange symmetry, so one uses a symmetric or antisymmetric subspace of the tensor product.

For mixed states, the corresponding joint-state description uses density operators on $\mathcal{H}_A \otimes \mathcal{H}_B$. The same distinction remains: the tensor product gives the joint quantum space, while Cartesian products are appropriate for pairing classical labels or subsystem descriptions.